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Which statements about inertia and centripetal force are correct? Check all that apply. Inertia is always present. Inertia cause

allsm [11]

Inertia IS always present. Inertia is NOT the force that causes objects to continue moving in circles, that is centripetal force. Centripetal force is NOT always present. Centripetal force DOES pull objects toward the center of a circle. <span> Inertia and centripetal force DOES cause circular motion. Thank you and eat sand fren ;)</span>

6 0

3 years ago

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Say you have a differential drive robot that has an axle length of 30cm and wheel diameter of 10cm. Find the angular velocity fo

Klio2033 [76]

Answer:

a) ω1 = 18rpm ω2 = -18rpm

b) ω1 = 102rpm ω2 = 138rpm

c) ω1 = ω2 = 3.18rpm

Explanation:

For the first case, we know that each wheel will spin in a different direction but with the same magnitude, so:

ωr = 6rpm This is the angular velocity of the robot

$\omega = \frac{\omega r * D/2}{r_{wheel}}$ where D is 30cm and rwheel is 5cm

$\omega = \frac{6 * 30/2}{5}=18rpm$ One velocity will be positive and the other will be negative:

ω1 = 18rpm ω2 = -18rpm

For part b, the formula is the same but distances change. Rcircle=100cm:

$\omega 1 = \frac{\omega r * (R_{circle} - D/2)}{r_{wheel}}$

$\omega 2 = \frac{\omega r * (R_{circle} + D/2)}{r_{wheel}}$

Replacing values, we get:

$\omega 1 = \frac{6 * (100 - 30/2)}{5}=102rpm$

$\omega 2 = \frac{\omega r * (100 + 30/2)}{5}=138rpm$

For part c, both wheels must have the same velocity:

$\omega = \frac{V_{robot}}{r_{wheel}}=20rad/min$

$\omega = 20rad/min * \frac{1rev}{2*\pi rad}=3.18rpm$

8 0

4 years ago

An interglacial is a period of time that occurs between Ice Ages. It’s marked by warmer temperatures and milder climates. During

Korolek [52]

Answer:

The correct answer is - 43%.

Explanation: The increase in CO2 between these two suggested periods is approximately 43%. Even though it is a natural process that the CO2 levels vary in the atmosphere, still this is not the same case nowadays. Nowadays, or rather in the past few decades, apart from the natural increase of CO2 in the atmosphere, it has seen a much more increased levels because of the human activity. The industrial facilities and the vehicles, the cutting of the forests and burning the wood (there's both release of CO2 from the burning of the trees and loss of natural accumulator of the CO2), are just some of the more important human activities that contribute to a significant rise in the CO2 levels.

3 0

3 years ago

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

Tom [10]

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d) E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

p_rad = I / c

Where I : Intensity = P / A

p_rad = P / A*c

Where A : Area of circular spot = pi*d^2 / 4

p_rad = 4P / pi*d^2*c

p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)

p_rad = 0.00408 Pa

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

λ_med = n_air*λ_air / n_med

λ_med = (1) * (810 nm) / 1.34

λ_med = 604 nm

f_med = f_air

f_med = c / λ_air

f_med = (3*10^8) / (810 * 10^-9)

f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d)

What is the electric and magnetic field amplitude in the laser beam?

I = P / A

I = 0.5*ε_o*c*E_o ^2

I = 4P / pi*d^2

Hence, E_o = ( 8 P / ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

E_o = 3.04 * 10^4 V / m

For maximum magnetic field strength:

B_o = E_o / c

B_o = 3.04 * 10^4 / (3*10^8)

B _o = 1.013 *10^-4 Nm/Amp

Answer: E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

5 0

3 years ago

((Psychology))An action that benefits others but not ourselves is an example of:

uranmaximum [27]

B altruism.
al·tru·ism
ˈaltro͞oˌizəm/Submit
noun
the belief in or practice of disinterested and selfless concern for the well-being of others

6 0

4 years ago

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